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Fixed formatting of README.md files.

Some typo fixes.

All entries work well.

Added some try commands.
This commit is contained in:
Cody Boone Ferguson committed 2023-03-18 10:03:51 -07:00
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commit fa2a763f6c
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1995/vanschnitz/README.md
+132 -130
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@@ -11,155 +11,157 @@
make all
### To run
## To run:
./vanschnitz
## Judges' comments
## Try:
To use:
make clobber all LEVEL=6
./vanschnitz
make clobber all LEVEL=8
## Judges' comments:
Warning: Values of LEVEL>8 have been known to cause compilers
(and in one case a system) to crash during compilation.
Warning: in the past, values of LEVEL>8 have been known to cause compilers (and
in one case a system) to crash during compilation. We tested it in 2023 on a
MacBook Pro with the M1 chip and found that LEVEL > 6 caused problems.
For a bad/slow time try LEVEL=15.
For a bad/slow time try LEVEL=15.
## Author's comments
## Authors' comments:
A classic problem in computer science is known as the Towers
of Hanoi. It involves a set of different-sized disks mounted
on one of three pegs. The object is to move the pile of disks
to one of the other pegs. You may only move one disk at a
time, and there is an additional constraint that a disk may
never be placed on top of a smaller disk.
A classic problem in computer science is known as the Towers
of Hanoi. It involves a set of different-sized disks mounted
on one of three pegs. The object is to move the pile of disks
to one of the other pegs. You may only move one disk at a
time, and there is an additional constraint that a disk may
never be placed on top of a smaller disk.
Our program solves the Towers of Hanoi problem. Well, that's
not exactly true; actually, it's the compiler that solves the
problem. The resulting program just prints out the correct
solution.
Our program solves the Towers of Hanoi problem. Well, that's
not exactly true; actually, it's the compiler that solves the
problem. The resulting program just prints out the correct
solution.
How do you trick a compiler into actually solving the problem?
First, you must tell it how many disks you wish to solve it
for. This is done by defining the symbol 'n' on the compile
line. For instance, to cause the compiler to solve the Towers
of Hanoi problem with four disks, you would compile the program
like this:
How do you trick a compiler into actually solving the problem?
First, you must tell it how many disks you wish to solve it
for. This is done by defining the symbol 'n' on the compile
line. For instance, to cause the compiler to solve the Towers
of Hanoi problem with four disks, you would compile the program
like this:
gcc hanoi.c -o hanoi -Dn=4
A default value of 5 will be used for n if you do not define
it on the command line. The value of n cannot be greater than
fifteen (the compiler we used to test has a limit on the #include
depth). The compiler then solves the problem using binary
arithmetic based on whether particular symbols are defined or not.
To loop, the program #include's itself. This is, of course,
expensive; one compile we did with n=14 took about fifty
minutes to compile on our system (compiling with n=15
caused our system to crash).
A default value of 5 will be used for n if you do not define
it on the command line. The value of n cannot be greater than
fifteen (the compiler we used to test has a limit on the #include
depth). The compiler then solves the problem using binary
arithmetic based on whether particular symbols are defined or not.
The resulting program that the compiler generates simply
prints out the answer. Did I say "simply"? Actually, the
whole resulting program consists of a single printf statement,
consisting of a massive string constant of length 35*(2^n-1),
followed by 3*(2^n-1) integers which get formatted into the
string. For our n=14 run, this adds up to a string constant
of length 573405, followed by 49149 integers delimeted by
commas. (Generating the string constant depends on the
ANSI C feature in which adjacent character strings are
catenated; a version that does not use this feature has been
included for people who can only run K&R). A good way to see
the resulting program (on a Unix system) is to do the command
To loop, the program #include's itself. This is, of course,
expensive; one compile we did with n=14 took about fifty
minutes to compile on our system (compiling with n=15
caused our system to crash).
The resulting program that the compiler generates simply
prints out the answer. Did I say "simply"? Actually, the
whole resulting program consists of a single printf statement,
consisting of a massive string constant of length `35*(2^n-1)`,
followed by `3*(2^n-1)` integers which get formatted into the
string. For our n=14 run, this adds up to a string constant
of length 573405, followed by 49149 integers delimited by
commas. (Generating the string constant depends on the
ANSI C feature in which adjacent character strings are
catenated; a version that does not use this feature has been
included for people who can only run K&R). A good way to see
the resulting program (on a Unix system) is to do the command
gcc hanoi.c -E -Dn=5 | grep -v \# | grep -v ^\$
For an odd number of disks, the program will provide a
solution wherein the disks end up on peg 2; for an even
number of disks, they will end on peg 3. This should
provide some hint as to what sort of algorithm is used.
For an odd number of disks, the program will provide a
solution wherein the disks end up on peg 2; for an even
number of disks, they will end on peg 3. This should
provide some hint as to what sort of algorithm is used.
We have included a "spoiler" version of the program, with
meaningful symbol names and comments, but we encourage you
to try to decipher the program without it...
We have included a "spoiler" version of the program, with
meaningful symbol names and comments, but we encourage you
to try to decipher the program without it...
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end
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